CS 61C: Great Ideas in
Computer Architecture
MIPS Instruction
Formats
Instructor: Justin Hsia
6/27/2012
Summer 2012 -- Lecture #7
1
Review of Last Lecture
• New registers: $a0-$a3, $v0-$v1, $ra, $sp
– Also: $at, $k0-k1, $gp, $fp, PC
• New instructions:
• Saved registers:
Volatile registers:
slt, la, li, jal, jr
$s0-$s7, $sp, $ra
$t0-$t9, $v0-$v1,
$a0-$a3
– CalleR saves volatile registers it is using before
making a procedure call
– CalleE saves saved registers it intends to use
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Summer 2012 -- Lecture #7
2
Question: Which statement below is TRUE about
converting the following C code to MIPS?
int factorial(int n) {
if(n == 0) return 1;
else return(n*factorial(n-1));
}
☐
This function MUST be implemented recursively
☐
We can write this function without using any
saved or temporary registers
We must save $ra on the stack since we
need to know where to return to
☐
☐
3
Great Idea #1: Levels of
Representation/Interpretation
temp = v[k];
v[k] = v[k+1];
v[k+1] = temp;
Higher-Level Language
Program (e.g. C)
Compiler
lw
lw
sw
sw
Assembly Language
Program (e.g. MIPS)
Assembler
Machine Language
Program (MIPS)
$t0, 0($2)
$t1, 4($2)
$t1, 0($2)
$t0, 4($2)
We__
are__
here._
0000 1001 1100 0110 1010 1111 0101 1000
1010 1111 0101 1000 0000 1001 1100 0110
1100 0110 1010 1111 0101 1000 0000 1001
0101 1000 0000 1001 1100 0110 1010 1111
Machine
Interpretation
Hardware Architecture Description
(e.g. block diagrams)
Architecture
Implementation
Logic Circuit Description
(Circuit Schematic Diagrams)
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Summer 2012 -- Lecture #7
4
Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Assembly Practice
• Bonus: Disassembly Practice
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Summer 2012 -- Lecture #7
5
Big Idea: Stored-Program Concept
• Encode your instructions as binary data
– Therefore, entire programs can be stored in
memory to be read or written just like data
• Simplifies SW/HW of computer systems
– Memory technology for data also used for
programs
• Stored in memory, so both instructions and
data words have addresses
– Use with jumps, branches, and loads
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Summer 2012 -- Lecture #7
6
Binary Compatibility
• Programs are distributed in binary form
– Programs bound to specific instruction set
– i.e. different versions for (old) Macs vs. PCs
• New machines want to run old programs (“binaries”)
as well as programs compiled to new instructions
• Leads to “backward compatible” instruction sets that
evolve over time
– The selection of Intel 80x86 in 1981 for 1st IBM PC is major
reason latest PCs still use 80x86 instruction set (Pentium
4); you could still run program from 1981 PC today
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Summer 2012 -- Lecture #7
7
Instructions as Numbers (1/2)
• Currently all data we work with is in words
(32-bit blocks)
– Each register is a word in length
– lw and sw both access one word of memory
• So how do we represent instructions?
– Remember: computer only understands 1s and
0s, so “add $t0,$0,$0” is meaningless.
– MIPS wants simplicity: since data is in words, let
instructions be in words, too
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Summer 2012 -- Lecture #7
8
Instructions as Numbers (2/2)
• Divide the 32 bits of an instruction into
“fields”
– Each field tells the processor something about the
instruction
– Could use different fields for every instruction, but
regularity leads to simplicity
• Define 3 types of instruction formats:
– R-Format
– I-Format
– J-Format
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Summer 2012 -- Lecture #7
9
Instruction Formats
• I-Format: instructions with immediates,
lw/sw (offset is immediate), and beq/bne
– But not the shift instructions
• J-Format: j and jal
– But not jr
• R-Format: all other instructions
• It will soon become clear why the instructions
have been partitioned in this way
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Summer 2012 -- Lecture #7
10
Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Assembly Practice
• Bonus: Disassembly Practice
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Summer 2012 -- Lecture #7
11
R-Format Instructions (1/4)
• Define “fields” of the following number of bits
each: 6 + 5 + 5 + 5 + 5 + 6 = 32
0
31
6
5
5
5
5
6
• For simplicity, each field has a name:
31
opcode
rs
rt
rd
0
shamt funct
• Each field is viewed as its own unsigned int
– 5-bit fields can represent any number 0-31,
while 6-bit fields can represent any number 0-63
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R-Format Instructions (2/4)
• opcode (6): partially specifies operation
– Set at 0b000000 for all R-Format instructions
• funct (6): combined with opcode, this number
exactly specifies the instruction
• How many R-format instructions can we encode?
– opcode is fixed, so 64
• Why aren’t these a single 12-bit field?
– We’ll answer this later
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Summer 2012 -- Lecture #7
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R-Format Instructions (3/4)
• rs (5): specifies register containing 1st operand
(“source register”)
• rt (5): specifies register containing 2nd operand
(“target register” – name is misleading)
• rd (5): specifies register that receives the result
of the computation (“destination register”)
• Recall: MIPS has 32 registers
– Fit perfectly in a 5-bit field (use register numbers)
• These map intuitively to instructions
– e.g. add dst,src1,src2  add rd,rs,rt
– Depending on instruction, field may not be used
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R-Format Instructions (4/4)
• shamt (5): The amount a shift instruction
will shift by
– Shifting a 32-bit word by more than 31 is useless
– This field is set to 0 in all but the shift instructions
• For a detailed description of field usage and
instruction type for each instruction, see the
MIPS Green Card
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Summer 2012 -- Lecture #7
15
R-Format Example (1/2)
• MIPS Instruction:
add
$8,$9,$10
• Pseudo-code (“OPERATION” column):
add
R[rd] = R[rs] + R[rt]
• Fields:
opcode = 0
funct = 32
rd = 8
rs = 9
rt = 10
shamt = 0
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(look up on Green Sheet)
(look up on Green Sheet)
(destination)
(first operand)
(second operand)
(not a shift)
Summer 2012 -- Lecture #7
16
R-Format Example (2/2)
• MIPS Instruction:
add $8,$9,$10
31
0
0
31
9
10
8
0
32
Field representation (binary):
0
000000 01001 01010 01000 00000 100000
two
hex representation:
0x 012A 4020
decimal representation: 19,546,144
Called a Machine Language Instruction
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NOP
• What is the instruction 0x00000000?
– opcode is 0, so is an R-Format
• Using Green Sheet, translates into:
sll $0,$0,0
– What does this do? Nothing!
• This is a special instruction called nop for “No
Operation Performed”
– We’ll see its uses later in the course
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Summer 2012 -- Lecture #7
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Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Converting to Machine Code Practice
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Summer 2012 -- Lecture #7
19
Administrivia
• HW 2 due Sunday
• Project 1 due 7/8
– No homework next week
– Will be released in the next two days
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Summer 2012 -- Lecture #7
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Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Assembly Practice
• Bonus: Disassembly Practice
6/27/2012
Summer 2012 -- Lecture #7
21
I-Format Instructions (1/4)
• What about instructions with immediates?
– 5- and 6-bit fields too small for most immediates
• Ideally, MIPS would have only one instruction
format (for simplicity)
– Unfortunately here we need to compromise
• Define new instruction format that is partially
consistent with R-Format
– First notice that, if instruction has immediate,
then it uses at most 2 registers
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22
I-Format Instructions (2/4)
• Define “fields” of the following number of bits
each: 6 + 5 + 5 + 16 = 32 bits
31
0
6
5
5
16
•
Field
names:
31
opcode
rs
0
rt
immediate
• Key Concept: Three fields are consistent with
R-Format instructions
– Most importantly, opcode is still in same location
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Summer 2012 -- Lecture #7
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I-Format Instructions (3/4)
• opcode (6): uniquely specifies the
instruction
– All I-Format instructions have non-zero opcode
– R-Format has two 6-bit fields to identify
instruction for consistency across formats
• rs (5): specifies a register operand
– Not always used
• rt (5): specifies register that receives result
of computation (“target register”)
– Name makes more sense for I-Format instructions
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Summer 2012 -- Lecture #7
24
I-Format Instructions (4/4)
• immediate (16): two’s complement number
– All computations done in words, so 16-bit
immediate must be extended to 32 bits
– Green Sheet specifies ZeroExtImm or SignExtImm
based on instruction
• Can represent 216 different immediates
– This is large enough to handle the offset in a
typical lw/sw, plus the vast majority of values for
slti
– We’ll see what to do when the number is too big
later today…
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Summer 2012 -- Lecture #7
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I-Format Example (1/2)
• MIPS Instruction:
addi
$21,$22,-50
• Pseudo-code (“OPERATION” column)
addi
R[rt] = R[rs] + SignExtImm
• Fields:
opcode = 8
rs = 22
rt = 21
immediate = -50
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(look up on Green Sheet)
(register containing operand)
(target register)
(decimal by default,
can also be specified in hex)
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26
I-Format Example (2/2)
• MIPS Instruction:
31
Field representation (decimal):
8
31
addi $21,$22,-50
22
21
0
-50
Field representation (binary):
001000 10110 10101
0
1111111111001110
two
hex representation:
0x 22D5 FFCE
decimal representation: 584,449,998
Called a Machine Language Instruction
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Summer 2012 -- Lecture #7
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Question: Which instruction has the same
representation as 35ten?
OPCODE/FUNCT:
subu 0/35
lw
35/-addi 8/--
Register names and numbers:
0: $0
8-15: $t0-$t7
16-23: $s0-$s7
☐
subu $s0,$s0,$s0 opcode
rs
rt
☐
lw
$0,0($0)
opcode
rs
rt
offset
☐
addi $0,$0,35
opcode
rs
rt
immediate
rd
shamt funct
☐
28
Dealing With Large Immediates
• How do we deal with 32-bit immediates?
– Sometimes want to use immediates > ± 215 with
addi, lw, sw and slti
– Bitwise logic operations with 32-bit immediates
• Solution: Don’t mess with instruction
formats, just add a new instruction
• Load Upper Immediate (lui)
– lui reg,imm
– Moves 16-bit imm into upper half (bits 16-31) of
reg and zeros the lower half (bits 0-15)
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lui Example
• Want: addi $t0,$t0,0xABABCDCD
– This is a pseudo-instruction!
• Translates into:
lui $at,0xABAB
# upper 16
ori $at,$at,0xCDCD # lower 16
add $t0,$t0,$at
# move
Only the assembler gets to use $at
• Now we can handle everything with a 16-bit
immediate!
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Branching Instructions
• beq and bne
– Need to specify an address to go to
– Also take two registers to compare
• Use I-Format:
31
opcode
rs
0
rt
immediate
– opcode specifies beq (4) vs. bne (5)
– rs and rt specify registers
– How to best use immediate to specify
addresses?
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Branching Instruction Usage
• Branches typically used for loops (if-else,
while, for)
– Loops are generally small (< 50 instructions)
– Function calls and unconditional jumps handled
with jump instructions (J-Format)
• Recall: Instructions stored in a localized area
of memory (Code/Text)
– Largest branch distance limited by size of code
– Address of current instruction stored in the
program counter (PC)
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PC-Relative Addressing
• PC-Relative Addressing: Use the immediate
field as a two’s complement offset to PC
– Branches generally change the PC by a small
amount
– Can specify ± 215 addresses from the PC
• So just how much of memory can we reach?
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Branching Reach
• Recall: MIPS uses 32-bit addresses
– Memory is byte-addressed
• Instructions are word-aligned
– Address is always multiple of 4 (in bytes), meaning it
ends with 0b00 in binary
– Number of bytes to add to the PC will always be a
multiple of 4
• Immediate specifies words instead of bytes
– Can now branch ± 215 words
– We can reach 216 instructions = 218 bytes around PC
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Branch Calculation
• If we don’t take the branch:
– PC = PC + 4 = next instruction
• If we do take the branch:
– PC = (PC+4) + (immediate*4)
• Observations:
– immediate is number of instructions to jump
(remember, specifies words) either forward (+) or
backwards (–)
– Branch from PC+4 for hardware reasons; will be
clear why later in the course
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Branch Example (1/2)
Start counting from
instruction AFTER the
branch
• MIPS Code:
Loop: beq
addu
addiu
j
End:
$9,$0,End
$8,$8,$10
$9,$9,-1
Loop
1
2
3
• I-Format fields:
opcode = 4
rs = 9
rt = 0
3
immediate = ???
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(look up on Green Sheet)
(first operand)
(second operand)
Summer 2012 -- Lecture #7
36
Branch Example (2/2)
• MIPS Code:
Loop: beq
addu
addiu
j
End:
31
$9,$0,End
$8,$8,$10
$9,$9,-1
Loop
Field representation (decimal):
4
9
0
0
3
31 Field representation (binary):
000100 01001 00000
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0
0000000000000011
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Questions on PC-addressing
• Does the value in branch immediate field
change if we move the code?
– If moving individual lines of code, then yes
– If moving all of code, then no
• What do we do if destination is > 215
instructions away from branch?
– Other instructions save us
– beq $s0,$0,far
# next instr
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bne $s0,$0,next
-->
j
far
next: # next instr
Summer 2012 -- Lecture #7
38
Get to Know Your Staff
• Category: Games
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39
Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Assembly Practice
• Bonus: Disassembly Practice
6/27/2012
Summer 2012 -- Lecture #7
40
J-Format Instructions (1/4)
• For branches, we assumed that we won’t want
to branch too far, so we can specify a change
in the PC
• For general jumps (j and jal), we may jump
to anywhere in memory
– Ideally, we would specify a 32-bit memory address
to jump to
– Unfortunately, we can’t fit both a 6-bit opcode
and a 32-bit address into a single 32-bit word
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J-Format Instructions (2/4)
• Define two “fields” of these bit widths:
31
6
26
• As usual, each field has a name:
31
opcode
0
0
target address
• Key Concepts:
– Keep opcode field identical to R-Format and
I-Format for consistency
– Collapse all other fields to make room for large
6/27/2012 target address
Summer 2012 -- Lecture #7
42
J-Format Instructions (3/4)
• We can specify 226 addresses
– Still going to word-aligned instructions, so add 0b00
as last two bits (multiply by 4)
– This brings us to 28 bits of a 32-bit address
• Take the 4 highest order bits from the PC
– Cannot reach everywhere, but adequate almost all of
the time, since programs aren’t that long
– Only problematic if code straddles a 256MB boundary
• If necessary, use 2 jumps or jr (R-Format)
instead
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J-Format Instructions (4/4)
• Jump instruction:
– New PC = { (PC+4)[31..28], target address, 00 }
• Notes:
– { , , } means concatenation
{ 4 bits , 26 bits , 2 bits } = 32 bit address
• Book uses || instead
– Array indexing: [31..28] means highest 4 bits
– For hardware reasons, use PC+4 instead of PC
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Question: When combining two C files into one
executable, we can compile them independently
and then merge them together.
When merging two or more binaries:
1) Jump instructions don’t require any changes
2) Branch instructions don’t require any changes
☐
☐
☐
1
F
F
T
2
F
T
F
☐
45
Summary
• The Stored Program concept is very powerful
– Instructions can be treated and manipulated the
same way as data in both hardware and software
•
R:
I:
J:
•
MIPS Machine Language Instructions:
opcode
rs
rt
opcode
rs
rt
opcode
rd
shamt funct
immediate
target address
Branches use PC-relative addressing,
Jumps use absolute addressing
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You are responsible for the material contained
on the following slides, though we may not have
enough time to get to them in lecture.
They have been prepared in a way that should
be easily readable and the material will be
touched upon in the following lecture.
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Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Assembly Practice
• Bonus: Disassembly Practice
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48
Assembly Practice
• Assembly is the process of converting assembly
instructions into machine code
• On the following slides, there are 6-lines of
assembly code, along with space for the machine
code
• For each instruction,
1)
2)
3)
4)
5)
Identify the instruction type (R/I/J)
Break the space into the proper fields
Write field values in decimal
Convert fields to binary
Write out the machine code in hex
• Use your Green Sheet; answers follow
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Code Questions
Addr
800
804
808
812
816
820
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Material from past lectures:
Instruction
Loop: sll $t1,$s3,2 What type of C variable is
addu
$t1,$t1,$s6
probably stored in $s6?
int * (or any pointer)
Write an equivalent C loop using
a$s3, b$s5, c$s6. Define
lw
$t0,0($t1)
variable types (assume they are
initialized somewhere) and feel
beq
$t0,$s5, Exit free to introduce other variables
as you like.
int a,b,*c;
addiu $s3,$s3,1
/* values initialized */
while(c[a] != b) a++;
j
Exit:
Loop
In English, what does this loop do?
Finds an entry in array c that
matches b.
Summer 2012 -- Lecture #7
50
Assembly Practice Question
Addr
800
Instruction
Loop: sll $t1,$s3,2
__:
804
addu
$t1,$t1,$s6
lw
$t0,0($t1)
beq
$t0,$s5, Exit
__:
808
__:
812
__:
816
addiu $s3,$s3,1
__:
820
j
Loop
__:
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Exit:
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Assembly Practice Answer (1/4)
Addr Instruction
800 Loop: sll $t1,$s3,2
R: opcode
rs
rt
rd
shamt funct
804 addu $t1,$t1,$s6
R: opcode
rs
rt
rd
shamt funct
808 lw
$t0,0($t1)
I: opcode
rs
rt
immediate
812 beq
$t0,$s5, Exit
I: opcode
rs
rt
immediate
816 addiu $s3,$s3,1
I: opcode
rs
rt
immediate
820 j
Loop
J: opcode
target address
Exit:
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Assembly Practice Answer (2/4)
Addr
800
R:
804
R:
808
I:
812
I:
816
I:
820
J:
6/27/2012
Instruction
Loop: sll $t1,$s3,2
0
0
19
addu $t1,$t1,$s6
0
9
22
lw
$t0,0($t1)
35
9
8
beq
$t0,$s5, Exit
4
8
21
addiu $s3,$s3,1
8
19
19
j
Loop
2
Exit:
Summer 2012 -- Lecture #7
9
2
0
9
0
33
0
2
1
200
53
Assembly Practice Answer (3/4)
Addr Instruction
800 Loop: sll $t1,$s3,2
R: 000000 00000 10011 01001 00010 000000
804 addu $t1,$t1,$s6
R: 000000 01001 10110 01001 00000 100001
808 lw
$t0,0($t1)
I: 100011 01001 01000 0000 0000 0000 0000
812 beq
$t0,$s5, Exit
I: 000100 01000 10101 0000 0000 0000 0010
816 addiu $s3,$s3,1
I: 001000 10011 10011 0000 0000 0000 0001
820 j
Loop
J: 000010 00 0000 0000 0000 0000 1100 1000
Exit:
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Assembly Practice Answer (4/4)
Addr
800
R:
804
R:
808
I:
812
I:
816
I:
820
J:
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Instruction
Loop: sll $t1,$s3,2
0x 0013 4880
addu $t1,$t1,$s6
0x 0136 4821
lw
$t0,0($t1)
0x 8D28 0000
beq
$t0,$s5, Exit
0x 1115 0002
addiu $s3,$s3,1
0x 2273 0001
j
Loop
0x 0800 00C8
Exit:
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Agenda
•
•
•
•
Stored-Program Concept
R-Format
Administrivia
I-Format
– Branching and PC-Relative Addressing
• J-Format
• Bonus: Assembly Practice
• Bonus: Disassembly Practice
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Disassembly Practice
• Disassembly is the opposite process of figuring
out the instructions from the machine code
• On the following slides, there are 6-lines of
machine code (hex numbers)
• Your task:
1)
2)
3)
4)
5)
Convert to binary
Use opcode to determine format and fields
Write field values in decimal
Convert fields MIPS instructions (try adding labels)
Translate into C (be creative!)
• Use your Green Sheet; answers follow
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Disassembly Practice Question
Address
0x00400000
...
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Instruction
0x00001025
0x0005402A
0x11000003
0x00441020
0x20A5FFFF
0x08100001
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Disassembly Practice Answer (1/9)
Address
0x00400000
...
Instruction
00000000000000000001000000100101
00000000000001010100000000101010
00010001000000000000000000000011
00000000010001000001000000100000
00100000101001011111111111111111
00001000000100000000000000000001
1) Converted to binary
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Disassembly Practice Answer (2/9)
Address
Instruction
0x00400000 R 00000000000000000001000000100101
R 00000000000001010100000000101010
...
I 00010001000000000000000000000011
R 00000000010001000001000000100000
I 00100000101001011111111111111111
J 00001000000100000000000000000001
2) Check opcode for format and fields...
– 0 (R-Format), 2 or 3 (J-Format), otherwise (I-Format)
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Disassembly Practice Answer (3/9)
Address
Instruction
0
0x00400000 R
0
R
...
4
I
0
R
8
I
2
J
0
0
8
2
5
0
5
0
4
5
2
8
0
0
+3
2
0
-1
0x0100001
37
42
32
3) Convert to decimal
– Can leave target address in hex
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Disassembly Practice Answer (4/9)
Address
0x00400000
0x00400004
0x00400008
0x0040000C
0x00400010
0x00400014
0x00400018
Instruction
or
$2,$0,$0
slt $8,$0,$5
beq $8,$0,3
add $2,$2,$4
addi $5,$5,-1
j
0x0100001
4) Translate to MIPS instructions (write in addrs)
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Summer 2012 -- Lecture #7
62
Disassembly Practice Answer (5/9)
Address
0x00400000
0x00400004
0x00400008
0x0040000C
0x00400010
0x00400014
0x00400018
Instruction
or
$v0,$0,$0
slt $t0,$0,$a1
beq $t0,$0,3
add $v0,$v0,$a0
addi $a1,$a1,-1
j
0x0100001 # addr: 0x0400004
4) Translate to MIPS instructions (write in addrs)
– More readable with register names
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Summer 2012 -- Lecture #7
63
Disassembly Practice Answer (6/9)
Address
0x00400000
0x00400004
0x00400008
0x0040000C
0x00400010
0x00400014
0x00400018
Instruction
or
Loop: slt
beq
add
addi
j
Exit:
$v0,$0,$0
$t0,$0,$a1
$t0,$0,Exit
$v0,$v0,$a0
$a1,$a1,-1
Loop
4) Translate to MIPS instructions (write in addrs)
– Introduce labels
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Summer 2012 -- Lecture #7
64
Disassembly Practice Answer (7/9)
Address
Instruction
or
$v0,$0,$0
Loop: slt $t0,$0,$a1
beq $t0,$0,Exit
add $v0,$v0,$a0
addi $a1,$a1,-1
j
Loop
Exit:
#
#
#
#
#
initialize $v0 to 0
$t0 = 0 if 0 >= $a1
exit if $a1 <= 0
$v0 += $a0
decrement $a1
4) Translate to MIPS instructions (write in addrs)
– What does it do?
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Summer 2012 -- Lecture #7
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Disassembly Practice Answer (8/9)
/* a$v0, b$a0, c$a1 */
a = 0;
while(c > 0) {
a += b;
c--;
}
5) Translate into C code
– Initial direct translation
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Summer 2012 -- Lecture #7
66
Disassembly Practice Answer (9/9)
/* naïve multiplication: returns m*n */
int multiply(int m, int n) {
int p; /* product */
for(p = 0; n-- > 0; p += m) ;
return p;
}
5) Translate into C code
– One of many possible ways to write this
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Summer 2012 -- Lecture #7
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